Problem: $A=\left[\begin{array}{rr}12 & -12 & 8 & 3 \\-1 & 8 & -2 & 3 \\1 &0 &-9 & 15 \\16 &12 &8 & 5 \\10 &14 &3 & 19\end{array}\right]$ $A_{4,1}=$
Explanation: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{4,1}$ $A_{{4},{1}}$ is located on row ${4}$ of $A$ : $\left[\begin{array}{rr}12 & -12 & 8 & 3 \\-1 & 8 & -2 & 3 \\1 &0 &-9 & 15 \\{16} &{12} &{8} & {5} \\10 &14 &3 & 19\end{array}\right]$ $A_{{4},{1}}$ is also located on column ${1}$ of $A$. $\left[\begin{array}{rr}{12} & -12 & 8 & 3 \\{-1} & 8 & -2 & 3 \\{1} &0 &-9 & 15 \\ {\text{16}} &{12} &{8} & {5} \\{10} &14 &3 & 19\end{array}\right]$ Therefore, $A_{{4},{1}}={16}$. Summary $A_{4,1}=16$